Last updated at Oct. 25, 2019 by Teachoo

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Ex 8.2, 2 Choose the correct option and justify your choice : (i) "2 tan 30 " /"1 + tan2 30 " = sin 60 (B) cos 60 (C) tan 60 (D) sin 30 We know that , tan "30 " = 1/ 3 So, (2 tan 30 )/(1+ 2 30 ) = (2 (1/ 3))/(1 + (1/ 3)^2 ) = (2/ 3)/(1 + 1/3) = (2/ 3)/(4/3) = 2/ 3 3/4 = 2/ 3 ( 3 3)/4 = 3/2 Options are (A) sin 60 (B) cos 60 (C) tan 60 (D) sin 30 Since sin "60 " = 3/2 Hence, option (A) is correct Ex 8.2, 2 Choose the correct option and justify your choice : (ii) "1 tan2 45 " /"1 + tan2 45 " = tan 90 (B) 1 (C) sin 45 (D) 0 We know that, tan 45 = 1 So, (1 tan ^2 45 )/(1 + tan ^2 45 ) = (1 (1)^2)/(1 + (1)^2 ) = (1 1)/(1 + 1) = 0/2 = 0 Hence, Option (D) is correct Ex 8.2, 2 Choose the correct option and justify your choice : (iii) sin 2A = 2 sin A is true when A = 0 (B) 30 (C) 45 (D) 60 sin 2A = 2 sin A Here, we substitute the value of option in question and whichever satisfies the question would be solution . Option (A) = 0 Lets s check it sin 2A = 2sin A sin 2(0) = 2 sin(0) sin 0 = 2 sin 0 0 = 0 L.H.S = R.H.S Hence option A is correct Option (B) = 30 sin 2A = 2 sin A sin 2(30 ) = 2(sin 30 ) sin 60 = 2 sin 30 3/2=2 1/2 3/2=1 But 3/2 " " 1 So, Option (B) is not possible Option (c) = 45 sin 2A = 2 sin A sin 2 (45 ) = 2 sin 45 sin 90 = 2 sin45 1 = 2 1/ 2 1 = 2 2 1/ 2 1 = 2 But 1 2 So, Option (C) cannot be possible Option (D) = 60 sin 2A = 2 sin A sin 2(60 ) = 2 sin(60 ) sin 120 = 2 sin 60 sin (90 + 30) = 2 sin 60 cos 30 = 2sin 60 3/2=2 3/2 3/2= 3 But 3/2 3 Hence, option (D) is incorrect So, only option (A) is correct Ex 8.2, 2 Choose the correct option and justify your choice : (iv) (2 π‘ππβ‘γ30Β°γ)/(1 β tan^2β‘γ30Β°γ ) (A) cos 60Β° (B) sin 60Β° (C) tan 60Β° (D) sin 30Β° We know that , tan "30Β°" = 1/β3 So, (2 tanβ‘γ30Β°γ)/(1 β π‘ππ2 30Β°) = (2 Γ (1/β3))/(1 β (1/β3)^2 ) = (2/β3)/(1 β 1/3) = (2/β3)/(2/3) = 2/β3Γ3/2 = 2/β3Γ(β3 Γ β3)/2 = β3 Options are (A) cos 60Β° (B) sin 60Β° (C) tan 60Β° (D) sin 30Β° Since tan "60Β°" =β3 Hence, option (C) is correct

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.